Let $X_1 \sim N(0,1)$ and $\xi$ be an independent symmetric random sign, i.e. $P(\xi = \pm1) = \frac{1}{2}$ Define $X_2= \xi X_1$. This random variable is normal, since for any $s\in\mathbb{R}$ (???)
Show that $X_1$ and $X_2$ are not independent.
I’ve no idea how to solve it, any suggestions?
Note that independence means $P(X_1<0, X_2 < 0) = P(X_1 < 0) P(X_2 < 0)$. We will prove that this is not true, hence, they are dependent.
$$P(X_2 < 0)=P(\xi X_1 < 0)=P(X_1<0, \xi >0) + P(X_1>0, \xi <0)=$$ $$P(X_1<0)P(\xi >0) + P(X_1>0)P(\xi <0) = \frac{1}{2}\cdot\frac{1}{2}+ \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$
But,
$$P(X_1<0, X_2<0)=P(X_1<0, \xi X_1 < 0) = P(X_1 < 0, \xi > 0)=$$ $$=P(X_1 < 0)P(\xi > 0)=1/4$$
Therefore, $P(X_1<0, X_2<0) = 1/4\neq P(X_1<0)P(X_2 <0) = 1/2 \cdot 1/4$.